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3 years ago

Why is Pluto no longer a planet?

Physics

2 answers:

Tema [17]3 years ago

7 0

Answer:

Scientist thought it was to small to be a planet.

Explanation:

it's sad, it was my favorite planet. now it's my favorite dwarf planet.

Step2247 [10]3 years ago

5 0

The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects.

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A 1423-kg car is moving along a level highway with a speed of 26.4 m/s. The driver takes the foot off the accelerator and the ca

Leona [35]

Answer:

Final speed, v = 28.81 m/s

Explanation:

Given that,

Mass of the car, m = 1423 kg

Initial speed of the car, u = 26.4 m/s

Force experience by the car, F = 901 N

Distance, d = 106 m

To find,

The speed of the car after traveling this distance.

Solution,

The force experienced by a car is equal to the product of mass and acceleration.

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{901}{1423}$

$a=0.63\ m/s^2$

Let v is the final speed of the car. Using third equation of motion to find it as :

$v^2-u^2=2ad$

$v^2=2ad+u^2$

$v^2=2\times 0.63\times 106+(26.4)^2$

v = 28.81 m/s

So, the final speed of the car is 28.81 m/s.

5 0

4 years ago

If the weight of the bowling ball acts down with a force of 200 N, what force would the table need to push up with to keep the b

zavuch27 [327]

5858585 8 8 855858 858 585858

3 0

4 years ago

How many times Father is Proxima Centauri from the sun then is earth from the sun answer choices about 269,000 times or about 0.

Bas_tet [7]

Answer:

269,000

Explanation:

The distance between Proxima Centauri and the Sun, d₁₁ = 4.246 light years

The distance between Earth and the Sun, d₂ = 1.58 × 10⁻⁵ light years

The ratio of the distances are;

d₁/d₂ = 4.246/(1.58 × 10⁻⁵) = 268734.177215 ≈ 269,000 times

Therefore, Proxima Centauri is approximately 269,000 times further from the Sun than the Earth is from the Sun

4 0

4 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.44 kg and rotate with

Free_Kalibri [48]

Answer:

(a) 20,154.1 J

(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

$I= \frac{1}{2}mr^{2}$

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

$K= \frac{1}{2}I\omega ^{2}$

Here, K is the rotational kinetic energy and $\omega$ is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.356 m.

$I= \frac{1}{2}(3.44)(0.356)^{2}$

$I= 0.218 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

$K= \frac{1}{2}I\omega ^{2}$

Put $I= 0.218 kg m^{2}$ and $\omega = 430 rads^{-1}$ .

$K= \frac{1}{2}(0.218)(430) ^{2}$

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.775 m.

$I= \frac{1}{2}(3.44)(0.775)^{2}$

$I= 1.03 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

$K= \frac{1}{2}I\omega ^{2}$

Put $I= 1.03 kg m^{2}$ and $\omega = 430 rads^{-1}$ .

$K= \frac{1}{2}(1.03)(430) ^{2}$

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.

7 0

4 years ago

Decide which examples are Nuclear Fission reaction.

MrMuchimi

a,c,d, is your answer I think I hope it helps

4 0

3 years ago

Read 2 more answers